How to get a uniform distribution in a range [r1,r2] in PyTorch?

Question

I want to get a 2-D torch.Tensor with size [a,b] filled with values from a uniform distribution (in range [r1,r2]) in PyTorch.

idk if it's just me but I find that `torch.rand` is very bad naming. I'd have no idea if it's Gaussian or Uniform without looking at the docs (that btw take ages to load for some reason on a browser). — Charlie Parker, Jul 13 '20 at 16:08
scratch my previous comment. Use this: `r2 = torch.torch.distributions.Uniform(low=lb, high=ub).sample((num_samples,Din)) ` — Charlie Parker, Jul 15 '20 at 16:44

score 87 · Accepted Answer · edited Jan 05 '20 at 22:14

87

If U is a random variable uniformly distributed on [0, 1], then (r1 - r2) * U + r2 is uniformly distributed on [r1, r2].

Thus, you just need:

(r1 - r2) * torch.rand(a, b) + r2

Alternatively, you can simply use:

torch.FloatTensor(a, b).uniform_(r1, r2)

To fully explain this formulation, let's look at some concrete numbers:

r1 = 2 # Create uniform random numbers in half-open interval [2.0, 5.0)
r2 = 5

a = 1  # Create tensor shape 1 x 7
b = 7

We can break down the expression (r1 - r2) * torch.rand(a, b) + r2 as follows:

torch.rand(a, b) produces an a x b (1x7) tensor with numbers uniformly distributed in the range [0.0, 1.0).

x = torch.rand(a, b)
print(x)
# tensor([[0.5671, 0.9814, 0.8324, 0.0241, 0.2072, 0.6192, 0.4704]])

(r1 - r2) * torch.rand(a, b) produces numbers distributed in the uniform range [0.0, -3.0)

print((r1 - r2) * x)
tensor([[-1.7014, -2.9441, -2.4972, -0.0722, -0.6216, -1.8577, -1.4112]])

(r1 - r2) * torch.rand(a, b) + r2 produces numbers in the uniform range [5.0, 2.0)

print((r1 - r2) * x + r2)
tensor([[3.2986, 2.0559, 2.5028, 4.9278, 4.3784, 3.1423, 3.5888]])

Now, let's break down the answer suggested by @Jonasson: (r2 - r1) * torch.rand(a, b) + r1

Again, torch.rand(a, b) produces (1x7) numbers uniformly distributed in the range [0.0, 1.0).

x = torch.rand(a, b)
print(x)
# tensor([[0.5671, 0.9814, 0.8324, 0.0241, 0.2072, 0.6192, 0.4704]])

(r2 - r1) * torch.rand(a, b) produces numbers uniformly distributed in the range [0.0, 3.0).

print((r2 - r1) * x)
# tensor([[1.7014, 2.9441, 2.4972, 0.0722, 0.6216, 1.8577, 1.4112]])

(r2 - r1) * torch.rand(a, b) + r1 produces numbers uniformly distributed in the range [2.0, 5.0)

print((r2 - r1) * x + r1)
tensor([[3.7014, 4.9441, 4.4972, 2.0722, 2.6216, 3.8577, 3.4112]])

In summary, (r1 - r2) * torch.rand(a, b) + r2 produces numbers in the range [r2, r1), while (r2 - r1) * torch.rand(a, b) + r1 produces numbers in the range [r1, r2).

edited Jan 05 '20 at 22:14

stackoverflowuser2010

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answered Jun 05 '17 at 18:56

BiBi

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shouldn't it be the other way round? I.e. (r2-r1)*torch.rand(a,b) + r1? – Jonasson Nov 05 '18 at 12:11
1

Isn’t it the same? – BiBi Nov 05 '18 at 12:32
@Jonasson: I think your answer is the better one. – stackoverflowuser2010 Jan 05 '20 at 22:15
@BiBi: No, they are not the same. I expanded your answer to give a more thorough explanation. Your answer produces numbers in the range `[r2, r1)`, while @Jonasson's answer produces numbers in the range `[r1, r2)`. Note that the intervals are half-open, meaning that your answer will never produce `r1` as a value, while @Jonasson's answer will. – stackoverflowuser2010 Jan 05 '20 at 22:18
I still don't understand why there isn't something like torch.uniform like there is is for Numpy. – Charlie Parker Jul 13 '20 at 16:06
idk if it's just me but I find that `torch.rand` is very bad naming. I'd have no idea if it's Gaussian or Uniform without looking at the docs (that btw take ages to load for some reason on a browser). – Charlie Parker Jul 13 '20 at 16:08
1

@stackoverflowuser2010 any single value in a continuous distribution has zero probability, so PDF's defined on a closed, open, or half open interval are [all equivalent](https://stats.stackexchange.com/questions/140805/in-the-definition-of-probability-density-function-does-it-matter-if-the-interva). – iacob Mar 12 '21 at 12:39
1

@CharlieParker [there is:](https://stackoverflow.com/a/66600118/9067615) `torch.distributions.uniform.Uniform()` – iacob Mar 12 '21 at 12:39
@iacob : Look at the APIs, not the math. – stackoverflowuser2010 Mar 12 '21 at 16:54

score 31 · Answer 2 · answered Dec 01 '17 at 23:08

31

torch.FloatTensor(a, b).uniform_(r1, r2)

answered Dec 01 '17 at 23:08

rafaelvalle

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While the accepted answer goes into more detail on different methods, and how they work, this answer is the simplest. – Joseph Summerhays Mar 07 '20 at 22:06
1

I still don't understand why there isn't something like torch.uniform like there is is for Numpy. – Charlie Parker Jul 13 '20 at 16:07
2

@CharlieParker [there is:](https://stackoverflow.com/a/66600118/9067615) `torch.distributions.uniform.Uniform() ` – iacob Mar 12 '21 at 12:43

Gaurav Shrivastava · Answer 3 · 2021-03-13T02:09:25.103

9

Utilize the torch.distributions package to generate samples from different distributions.

For example to sample a 2d PyTorch tensor of size [a,b] from a uniform distribution of range(low, high) try the following sample code

import torch
a,b = 2,3   #dimension of the pytorch tensor to be generated
low,high = 0,1 #range of uniform distribution

x = torch.distributions.uniform.Uniform(low,high).sample([a,b])

edited Mar 13 '21 at 02:09

answered Nov 18 '20 at 23:08

Gaurav Shrivastava

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score 3 · Answer 4 · edited Apr 04 '21 at 12:20

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Please Can you try something like:

import torch as pt
pt.empty(2,3).uniform_(5,10).type(pt.FloatTensor)

edited Apr 04 '21 at 12:20

Machavity

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answered May 28 '19 at 09:49

Sunil

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iacob · Answer 5 · 2021-03-12T15:13:42.703

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PyTorch has a number of distributions built in. You can build a tensor of the desired shape with elements drawn from a uniform distribution like so:

from torch.distributions.uniform import Uniform

shape = 3,4
r1, r2 = 0,1

x = Uniform(r1, r2).sample(shape)

edited Mar 12 '21 at 15:13

answered Mar 12 '21 at 12:41

iacob

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score 2 · Answer 6 · answered Oct 19 '18 at 18:42

To get a uniform random distribution, you can use

torch.distributions.uniform.Uniform()

example,

import torch
from torch.distributions import uniform

distribution = uniform.Uniform(torch.Tensor([0.0]),torch.Tensor([5.0]))
distribution.sample(torch.Size([2,3])

This will give the output, tensor of size [2, 3].

stackoverflowuser2010 · Answer 7 · 2021-03-13T17:12:41.817

2

This answer uses NumPy to first produce a random matrix and then converts the matrix to a PyTorch tensor. I find the NumPy API to be easier to understand.

import numpy as np

torch.from_numpy(np.random.uniform(low=r1, high=r2, size=(a, b)))

edited Mar 13 '21 at 17:12

answered Jan 05 '20 at 22:23

stackoverflowuser2010

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score 0 · Answer 8 · edited Apr 04 '21 at 12:20

See this for all distributions: https://pytorch.org/docs/stable/distributions.html#torch.distributions.uniform.Uniform

This is the way I found works:

# generating uniform variables

import numpy as np

num_samples = 3
Din = 1
lb, ub = -1, 1

xn = np.random.uniform(low=lb, high=ub, size=(num_samples,Din))
print(xn)

import torch

sampler = torch.distributions.Uniform(low=lb, high=ub)
r = sampler.sample((num_samples,Din))

print(r)

r2 = torch.torch.distributions.Uniform(low=lb, high=ub).sample((num_samples,Din))

print(r2)

# process input
f = nn.Sequential(OrderedDict([
    ('f1', nn.Linear(Din,Dout)),
    ('out', nn.SELU())
]))
Y = f(r2)
print(Y)

but I have to admit I don't know what the point of generating sampler is and why not just call it directly as I do in the one liner (last line of code).

Comments:

sampler are good for it's so you can transform/compose/cache/etc distributions. see https://arxiv.org/abs/1711.10604, and the top of the docs of https://pytorch.org/docs/stable/distributions.html# and https://arxiv.org/abs/1506.05254
you can feed in tensors to uniform to let it know the high dimensional interval (hypercube) to generate the uniform samples (that's why it receives tensors as input rather than simply numbers)

Reference:

score 0 · Answer 9 · answered Jul 11 '22 at 07:36

Pytorch (now?) has a random integer function that allows:

torch.randint(low=r1, high=r2, size=(1,), **kwargs)

and returns uniformly sampled random integers of shape size in range [r1, r2).

https://pytorch.org/docs/stable/generated/torch.randint.html

How to get a uniform distribution in a range [r1,r2] in PyTorch?

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